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Let $u,v,w \in V$ a vector space over a field F such that $u \neq v \neq w$. If $\{ u , v , w \}$ is a basis for $V$, then prove that $\{ u+v+w , v+w , w \}$ is also a basis for $V$.

Proof:

Let $u,v,w \in V$ a vector space over a field $F$ such that $u \neq v \neq w$. Let $\{ u , v , w \}$ be a basis for $V$. Because $\{ u , v , w \}$ is a basis, then $u,v,w$ are linearly independent and $ \langle \{ u , v , w \} \rangle = V$.

Let $x \in V$ be an arbitrary vector then $x$ can be uniquely expressed as a linear combination of $\{ u , v , w \}$. Let's suppose $x=au+bv+cw$ for some $a,b,c \in F$.

On the other hand, let us consider $\{ u+v+w , v+w , w \} \subseteq V$.

Then $$ \begin{align*} \langle \{ u+v+w , v+w , w \} \rangle &= \{d(u+v+w) + e(v+w) + f(w) \mid d,e,f \in F\} \\ &= \{du + (d+e)v +(d+e+f)w \mid d,e,f \in F \} . \end{align*}$$

If $x \in V$, then $x=du + (d+e)v +(d+e+f)w$ is another unique representation of $x \in V$ . Then for any arbitrary $x \in V$, we have $d=a$, $d+e=b $and $d+e+f=c \in F$.

Because $\{ u , v , w \}$ is a basis for $V$, then $\{ u+v+w , v+w , w \}$ must also be a basis for $V$.

Srivatsan
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aortizmena
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  • I tried to give an alternate proof instead of proving <{ u , v , w }>=V<{ u+v+w , v+w , w }> – aortizmena Dec 06 '11 at 05:39
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    Note that $u\neq v\neq w$ only means that $u\neq v$ and $v\neq w$, and does not mean that also $u\neq w$ ("is not equal to" is not transitive). Note also that saying that the vectors are unequal is redundant when you assume they form a basis. No two elements of a linearly independent set can be equal. – Jonas Meyer Dec 06 '11 at 05:41
  • Thanks. I typed that because that is how the statement was given. I do undestand that no two elements of a linearly independent set can be equal. – aortizmena Dec 06 '11 at 05:48
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    Are you aware of the fact that you can type LaTeX commands for your math notation? –  Dec 06 '11 at 05:49
  • I should say that you are very, very close to proving that they do span. You just need to take in an arbitrary triple $a, b, c$ and produce $d, e, f$ such that the equations in your penultimate paragraph hold. – Dylan Moreland Dec 06 '11 at 06:00
  • Isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ? My idea was exactly to take something in the span of {u+v+w,v+w,w} and writte it as a linear combination of my original basis elements with the above unique expresion for any x∈V – aortizmena Dec 06 '11 at 06:04
  • @AndresOrtizMena you say "My idea was exactly to take something in the span of ${u + v + w, v + w, w}$ and write it as a linear combination of my original basis elements" but you can always do this, simply because the original basis, is a basis. What you need to do is take something arbitrary in the span of your original basis (namely in $V$) and write it as a linear combination of your new set. – Deven Ware Dec 06 '11 at 06:19
  • @Andres I think we're just misinterpreting each other. I agree with your first sentence and I think you have a fine strategy; I just couldn't tell how you were concluding. I would add the assignment of $d, e, f$ to your proof. You should probably spell out why $d, e, f$ are uniquely determined by $x$, as well. – Dylan Moreland Dec 06 '11 at 06:21
  • because for any arbitrary x in V we have a unique representation of x with the original basis and we can make another unique representation of x as a linear combination of my new basis thus we can find those unique constants looking that the span of my new basis can be "stated" as a linear combination of the old one so the constants can be uniquely determined – aortizmena Dec 06 '11 at 06:31
  • Another proof, essentially equivalent to the others suggested, but using different technology, would be to write the relationship between the original base and the new combination in matrix form, and then write down the inverse. In more complex cases a non-zero determinant would mean an invertible transformation, and the rank of the matrix would give the dimension of the image. But I guess this technology comes later in the development of the subject. – Mark Bennet Dec 06 '11 at 07:12

3 Answers3

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In general I find it much harder to show that a set of vectors spans the vector space, than showing a set is independent. If you want to go about your approach and show they span, what you would need to do is take an arbitrary vector in $V$ and write it is a linear combination of your new set.

An alternative way is to show that they are independent, which turns out to be quite simple.

Suppose $$c_1(u + v + w) + c_2(v+w) + c_3(w) = 0$$ then we know that $$c_1u + (c_1 + c_2)v + (c_1 + c_2 + c_3) w = 0$$ but $u, v, w$ are independent and so $$c_1 = c_1 + c_2 = c_1 + c_2 + c_3 = 0$$ From here its pretty clear that $c_1 = c_2 = c_3 = 0$ which would prove the claim.

EDIT: Note, if we show that the vectors are independent they must also span $V$, because in any vector space of dimension $n$, if we have $n$ independent vectors, they must span the vector space. Similarly, if we were able to show that they spanned the space then they would have to be independent since we have $\dim(V)$ of them

Deven Ware
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It is easily seen that each of your original basis elements are in the span of $\{u+v+w,v+w,w\}$. Since $\{u, v, w\}$ spans $V$, so does $\{u+v+w,v+w,w\}$ (if $A\subseteq {\rm span} \,(V)$, then ${\rm span}(A)\subseteq{\rm span }(V) \thinspace $).

So, we have a set of three vectors that span a space of dimension three. It follows that the set is independent and thus a basis.

David Mitra
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The proof is essentially correct, but you do have some unnecessary details. Removing redundant information, we can reduce it to the following:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Any $x \in V$ can be uniquely written $x = au + bv + cw$ for some $a, b, c \in F$. Let $d, e, f \in F$ be the unique choices so that $d = a$, $e = b - a$, and $f = c - b$; then $x$ can be uniquely written as $x = du + (d + e)v + (d + e + f)w$. Rearranging, we have $x = d(u + v + w) + e(v + w) + fw$. Since this is a unique expression for an arbitrary $x \in V$, then $\{u + v + w, v + w, w\}$ is a basis for $V$.

There is also this nice alternate proof:

Let $V$ be a vector space over $F$ and $\{u, v, w\}$ a basis for $V$. Let's start a new basis with $w$. Since $\{u, v, w\}$ is linearly independent, $v + w \notin span\{w\}$ and $u + v + w \notin span\{w, v + w\}$, so $\{w, v + w, u + v + w\}$ is linearly independent. It is also the same length as a basis, so it must be a basis itself.

smackcrane
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  • I do not follow your proof, but I suspect that it is incomplete. You demonstrate that $x$ can be written as a linear combination of the vectors ${ u+v+w, v+w, v }$, and I follow this part. But I do not see where you prove the uniqueness of this representation. – Srivatsan Dec 06 '11 at 06:05
  • the uniqueness follows because the sets ${u+v+w, v+w, v}$ and ${u,v,w}$ have the same cardinality, no? –  Dec 06 '11 at 06:09
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    It all depends on the theorems you have, I guess. – Dylan Moreland Dec 06 '11 at 06:10
  • Isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ? – aortizmena Dec 06 '11 at 06:10
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    since $x = au + bv + cw$ is unique and each $(a, b, c)$ corresponds to exactly one $(d, e, f)$, then $x = d(u + v + w) + e(v + w) + fw$ must also be unique. – smackcrane Dec 06 '11 at 06:12